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Here $\mathbb{F}_{q}$ means a finite field with $q = p^m$ elements where $p$ is the characteristic of the field in question and $\overline{\mathbb{F}_{q}}$ means its algebraic closure.

I am studying an article about the existence of a normal basis with a primitive element for finite extensions of $\mathbb{F}_{q}$. And there is this new definition for me, that I haven't found anywhere:

Let $\alpha \in \overline{\mathbb{F}_{q}}$, and let $\sigma$ the Frobenius Authomorphism ($\sigma(\alpha) = \alpha^q$). One can turn $\overline{\mathbb{F}_{q}}$ into a $\mathbb{F}_{q}[X]$-module as follows:

If $f \in \mathbb{F}_{q}[X]$, write $f = \sum_{i=0}^{k} a_{i}X^{i}$ and then define:

$f \circ \alpha := \sum_{i=0}^{k} a_{i}\sigma^{i}(\alpha)$

Which is essentially $f(\sigma)(\alpha)$.

We then have that

$\alpha \in \mathbb{F}_{q^n} \Leftrightarrow \alpha^{q^n} = \alpha \Leftrightarrow \sigma^{n}(\alpha) = \alpha \Leftrightarrow (X^n-1)\circ\alpha =0$,

and then the Annihilator of $\alpha$ over $\mathbb{F}_{q}[X]$ is non-trivial for every $\alpha$. Then one can define the, let's call here additive order ($Ord(\alpha)$) of $\alpha$ by the monic polynomial generating the Annihilator of $\alpha$. It's easy to see that $\alpha \in \mathbb{F}_{q^n} \Leftrightarrow Ord(\alpha)|X^n-1$.

The point is, in the article, it is said that:

If $\alpha \in \mathbb{F}_{q^n}$ has $Ord(\alpha) = g$, then $\alpha = h\circ\beta$, for some $\beta \in \mathbb{F}_{q^n}$ and $h = \dfrac{X^n-1}{g}$.

The point is, I just cannot prove this (it seems trivial since every source I look into just states this and does not prove).

Well, if $n$ does not divide the characteristic of $\mathbb{F}_{q^n}$, then it is clear for me, since $X^n-1$ is separable over $\mathbb{F}_{q}[X]$ and then $(g,h) = 1$, therefore there exists $s,t \in \mathbb{F}_{q}[X]$ such that

$sg + th = 1$, which implies $(sg + th)\circ \alpha = 1\circ \alpha$ and $ (sg)\circ \alpha + (th)\circ\alpha = (th)\circ\alpha = h\circ(t\circ\alpha) = \alpha$, defining $\beta = t \circ \alpha$ we get the result. $(sg)\circ \alpha = 0$ because $g$ is the order of $\alpha$ therefore $g\circ\alpha = 0$, and I used the fact that we have a module structure over $\mathbb{F}_{q}[X]$.

Now, if $n$ divide the characteristic of $\mathbb{F}_{q^n}$, I cannot prove by this way. Well, I think there is a simpler argument such that one doesn't need to split the proof between two cases. Can anyone help me with the second case or can anyone give me a hint or a new argument that helps me to solve this problem?

Here $\mathbb{F}_{q}$ means a finite field with $q = p^m$ elements where $p$ is the characteristic of the field in question and $\overline{\mathbb{F}_{q}}$ means its algebraic closure.

I am studying an article about the existence of a normal basis with a primitive element for finite extensions of $\mathbb{F}_{q}$. And there is this new definition for me, that I haven't found anywhere:

Let $\alpha \in \overline{\mathbb{F}_{q}}$, and let $\sigma$ the Frobenius Authomorphism ($\sigma(\alpha) = \alpha^q$). One can turn $\overline{\mathbb{F}_{q}}$ into a $\mathbb{F}_{q}[X]$-module as follows:

If $f \in \mathbb{F}_{q}[X]$, write $f = \sum_{i=0}^{k} a_{i}X^{i}$ and then define:

$f \circ \alpha := \sum_{i=0}^{k} a_{i}\sigma^{i}(\alpha)$

Which is essentially $f(\sigma)(\alpha)$.

We then have that

$\alpha \in \mathbb{F}_{q^n} \Leftrightarrow \alpha^{q^n} = \alpha \Leftrightarrow \sigma^{n}(\alpha) = \alpha \Leftrightarrow (X^n-1)\circ\alpha =0$,

and then the Annihilator of $\alpha$ over $\mathbb{F}_{q}[X]$ is non-trivial for every $\alpha$. Then one can define the, let's call here additive order ($Ord(\alpha)$) of $\alpha$ by the monic polynomial generating the Annihilator of $\alpha$. It's easy to see that $\alpha \in \mathbb{F}_{q^n} \Leftrightarrow Ord(\alpha)|X^n-1$.

The point is, in the article, it is said that:

If $\alpha \in \mathbb{F}_{q^n}$ has $Ord(\alpha) = g$, then $\alpha = h\circ\beta$, for some $\beta \in \mathbb{F}_{q^n}$ and $h = \dfrac{X^n-1}{g}$.

The point is, I just cannot prove this (it seems trivial since every source I look into just states this and does not prove).

Well, if the characteristic of $\mathbb{F}_{q^n}$ does not divide $n$, then it is clear for me, since $X^n-1$ is separable over $\mathbb{F}_{q}[X]$ and then $(g,h) = 1$, therefore there exists $s,t \in \mathbb{F}_{q}[X]$ such that

$sg + th = 1$, which implies $(sg + th)\circ \alpha = 1\circ \alpha$ and $ (sg)\circ \alpha + (th)\circ\alpha = (th)\circ\alpha = h\circ(t\circ\alpha) = \alpha$, defining $\beta = t \circ \alpha$ we get the result. $(sg)\circ \alpha = 0$ because $g$ is the order of $\alpha$ therefore $g\circ\alpha = 0$, and I used the fact that we have a module structure over $\mathbb{F}_{q}[X]$.

Now, if $n$ divide the characteristic of $\mathbb{F}_{q^n}$, I cannot prove by this way. Well, I think there is a simpler argument such that one doesn't need to split the proof between two cases. Can anyone help me with the second case or can anyone give me a hint or a new argument that helps me to solve this problem?

Here $\mathbb{F}_{q}$ means a finite field with $q = p^m$ elements where $p$ is the characteristic of the field in question.

I am studying an article about the existence of a normal basis with a primitive element for finite extensions of $\mathbb{F}_{q}$. And there is this new definition for me, that I haven't found anywhere:

Let $\alpha \in \overline{\mathbb{F}_{q}}$, and let $\sigma$ the Frobenius Authomorphism ($\sigma(\alpha) = \alpha^q$). One can turn $\overline{\mathbb{F}_{q}}$ into a $\mathbb{F}_{q}[X]$-module as follows:

If $f \in \mathbb{F}_{q}[X]$, write $f = \sum_{i=0}^{n} a_{i}X^{i}$ and then define:

$f \circ \alpha := \sum_{i=0}^{n} a_{i}\sigma^{i}(\alpha)$

Which is essentially $f(\sigma)(\alpha)$.

We then have that

$\alpha \in \mathbb{F}_{q^n} \Leftrightarrow \alpha^{q^n} = \alpha \Leftrightarrow \sigma^{n}(\alpha) = \alpha \Leftrightarrow (X^n-1)\circ\alpha =0$,

and then the Annihilator of $\alpha$ over $\mathbb{F}_{q}[X]$ is non-trivial for every $\alpha$. Then one can define the, let's call here additive order ($Ord(\alpha)$) of $\alpha$ by the monic polynomial generating the Annihilator of $\alpha$. It's easy to see that $\alpha \in \mathbb{F}_{q^n} \Leftrightarrow Ord(\alpha)|X^n-1$.

The point is, in the article, it is said that:

If $\alpha \in \mathbb{F}_{q^n}$ has $Ord(\alpha) = g$, then $\alpha = h\circ\beta$, for some $\beta \in \mathbb{F}_{q^n}$ and $h = \dfrac{X^n-1}{g}$.

The point is, I just cannot prove this (it seems trivial since every source I look into just states this and does not prove).

Well, if $n$ does not divide the characteristic of $\mathbb{F}_{q^n}$, then it is clear for me, since $X^n-1$ is separable over $\mathbb{F}_{q}[X]$ and then $(g,h) = 1$, therefore there exists $s,t \in \mathbb{F}_{q}[X]$ such that

$sg + th = 1$, which implies $(sg + th)\circ \alpha = 1\circ \alpha$ and $ (sg)\circ \alpha + (th)\circ\alpha = (th)\circ\alpha = h\circ(t\circ\alpha) = \alpha$, defining $\beta = t \circ \alpha$ we get the result. $(sg)\circ \alpha = 0$ because $g$ is the order of $\alpha$ therefore $g\circ\alpha = 0$, and I used the fact that we have a module structure over $\mathbb{F}_{q}[X]$.

Now, if $n$ divide the characteristic of $\mathbb{F}_{q^n}$, I cannot prove by this way. Well, I think there is a simpler argument such that one doesn't need to split the proof between two cases. Can anyone help me with the second case or can anyone give me a hint or a new argument that helps me to solve this problem?

Here $\mathbb{F}_{q}$ means a finite field with $q = p^m$ elements where $p$ is the characteristic of the field in question and $\overline{\mathbb{F}_{q}}$ means its algebraic closure.

I am studying an article about the existence of a normal basis with a primitive element for finite extensions of $\mathbb{F}_{q}$. And there is this new definition for me, that I haven't found anywhere:

Let $\alpha \in \overline{\mathbb{F}_{q}}$, and let $\sigma$ the Frobenius Authomorphism ($\sigma(\alpha) = \alpha^q$). One can turn $\overline{\mathbb{F}_{q}}$ into a $\mathbb{F}_{q}[X]$-module as follows:

If $f \in \mathbb{F}_{q}[X]$, write $f = \sum_{i=0}^{k} a_{i}X^{i}$ and then define:

$f \circ \alpha := \sum_{i=0}^{k} a_{i}\sigma^{i}(\alpha)$

Which is essentially $f(\sigma)(\alpha)$.

We then have that

$\alpha \in \mathbb{F}_{q^n} \Leftrightarrow \alpha^{q^n} = \alpha \Leftrightarrow \sigma^{n}(\alpha) = \alpha \Leftrightarrow (X^n-1)\circ\alpha =0$,

and then the Annihilator of $\alpha$ over $\mathbb{F}_{q}[X]$ is non-trivial for every $\alpha$. Then one can define the, let's call here additive order ($Ord(\alpha)$) of $\alpha$ by the monic polynomial generating the Annihilator of $\alpha$. It's easy to see that $\alpha \in \mathbb{F}_{q^n} \Leftrightarrow Ord(\alpha)|X^n-1$.

The point is, in the article, it is said that:

If $\alpha \in \mathbb{F}_{q^n}$ has $Ord(\alpha) = g$, then $\alpha = h\circ\beta$, for some $\beta \in \mathbb{F}_{q^n}$ and $h = \dfrac{X^n-1}{g}$.

The point is, I just cannot prove this (it seems trivial since every source I look into just states this and does not prove).

Well, if $n$ does not divide the characteristic of $\mathbb{F}_{q^n}$, then it is clear for me, since $X^n-1$ is separable over $\mathbb{F}_{q}[X]$ and then $(g,h) = 1$, therefore there exists $s,t \in \mathbb{F}_{q}[X]$ such that

$sg + th = 1$, which implies $(sg + th)\circ \alpha = 1\circ \alpha$ and $ (sg)\circ \alpha + (th)\circ\alpha = (th)\circ\alpha = h\circ(t\circ\alpha) = \alpha$, defining $\beta = t \circ \alpha$ we get the result. $(sg)\circ \alpha = 0$ because $g$ is the order of $\alpha$ therefore $g\circ\alpha = 0$, and I used the fact that we have a module structure over $\mathbb{F}_{q}[X]$.

Now, if $n$ divide the characteristic of $\mathbb{F}_{q^n}$, I cannot prove by this way. Well, I think there is a simpler argument such that one doesn't need to split the proof between two cases. Can anyone help me with the second case or can anyone give me a hint or a new argument that helps me to solve this problem?